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SENIOR MATHEMATICS COMPETITION 2013

Preliminary round Thursday 16th May 2013 Time allowed 1 ½ hours

Instructions Attempt all questions. It is not expected that you will finish them all. Full working should accompany all solutions. Calculators may be used, but no other reference material is permitted. Total: 52 marks

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NZAMT

SENIOR MATHEMATICS COMPETITION 2013

  1. A function f(x) obeys the rule f(x+1) = f(x) – f(x-1) + 1. Given that f(2) = 8 and f(1) = 3, find f(2013). [5]

Pattern: 3,8,6,-1,-6,-4,3 repeats √√√then 2013/6 has rem 3 √so f(2013) = 6√

  1. For a Sierpinski Triangle (fractal: see diagram below) formed by successively bisecting sides of the shaded triangles, joining the bisection points and unshading the interior triangle, with S0 having no unshaded triangles, S1 having one unshaded triangle (and, so, four non-overlapping interior triangles), show that the number of non-overlapping interior triangles for Sn is given by (3n+1 – 1)/2. S1 has 4 = 3 + 1, S2 has 13 =32 + 3 + 1, S3 has 40 √sequence of powers of 3, plus 1. √Use Sum of geometric sequence,√ simplify to get result.√ [4]

Sn has 3n-1x4+[3n-2+3n-3+…+3+1]=3n-1x4+ sum of GP, viz [3n-1-1]/2 gives result

2. Prove that √2 is irrational. [3]

Let sqrt 2 = a/b, where a and b are mutually prime,√ square√ and get contradiction.√

4. A disease test has a sensitivity of 0.95 (i.e. of those who test positive, 95% have the disease) and a specificity of 0.95 (of those who test negative, 95% do not have the disease). (a) Show that if the true prevalence of the disease is 40% that over 90% of test results will be correct, and give the percentages of false positives and false negatives correct to 1d.p. Table,√ 7.3%, √3.4% √ [3] (b) Show that if the true prevalence is 2%, that it is less likely that a laboratory positive result is correct, and give the percentages of false positives and negatives correct to 1d.p. Table, √72.1%,√ 0.1% √ [3] (c) Make a generalisation about disease testing. Low prevalence implies less likely lab positive correct. √ [1]

[5]

  1. For two numbers x and y, the arithmetic mean is A=(x+y)/2, the geometric mean is G=√(xy) and the harmonic mean is H=2/(1/x + 1/y). Prove that √(AH/2) - √2 G + G/√2 = 0 for all values of x and y. [3] LHS=√(xy/2)-√2√(xy)+√(xy)/√2 √= simplifies √= RHS√
  1. Let A = sin(π – t) – cos(π – t) and B = sin(π + t) – cos(π + t) . Find an expression for the difference D = A - B in terms of sin(t) and give the maximum and minimum values this fluctuates between. [4] A=sint+cost, √B=cost-sint, √D=2sint; √values are -2 to 2. √


8. An equilateral triangle ACB is drawn, with arcs AEC and BDC each of radius AB. Find the ratio Area of semicircle on AB: Area AECDB. [3] Areas of sector, triangle, segment, semicircle; numerical or algebraic √√give 3 : 8 - 6√3/π. √Accept 3π: (8π - 6√3)

Q4, draw up a table with True: Pos, Neg, Total across the top; Test: Pos, Neg, Total down the side. For the 40% fill in left to right, top to bottom: 38, 3, 41; 2, 57, 59; 40, 60, 100 so then calculate false positives as 7.3% and false negatives as 3.4% For the 2% the figures will be: 1.9, 4.9, 6.8; 0.1, 93.1, 93.2; 2, 98, 100 and false pos 72.1%, false neg 0.1%.

However, I (now) realise that the comment in the brackets in the question is wrong and makes the question nonsensical. If the student uses Occam's razor and answers what the question means; the answer above is correct.

If they give some other answer, sorry, you might have to choose to give them credit for a sensible answer

Question 14: I see that I have left the "k greater than zero" off the question, so of course there are infinitely many values for k.